A red ball is thrown down with an initial speed of 1 m/s from a height of 26 meters above the ground. Then, 0.6 seconds after the red ball is thrown, a blue ball is thrown upward with an initial speed of 23.5 m/s, from a height of 1 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s2.
use your basic equation for falling bodies. The height
h(t) = h0 + v0t - 1/2 gt^2
How far do you get?
To determine the time it takes for the red ball to hit the ground, we can use the equation of motion:
s = ut + (1/2)at^2
where:
s = displacement (in this case, the height from which the ball is thrown, 26 meters)
u = initial velocity (1 m/s)
t = time
a = acceleration due to gravity (-9.81 m/s^2, negative because it acts in the opposite direction of the motion)
Plugging in the values, we get:
26 = 1t + (1/2)(-9.81)t^2
Simplifying the equation, we have:
-4.905t^2 + t + 26 = 0
This is a quadratic equation, which can be solved using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = -4.905, b = 1, and c = 26. Plugging in these values into the quadratic formula, we get:
t = (-1 ± √(1 - 4(-4.905)(26))) / (2(-4.905))
Simplifying further, we have:
t ≈ (-1 ± √(1 + 509.88)) / (-9.81)
Using a calculator, we find that the solutions are approximately t ≈ 0.947s and t ≈ 5.380s. Since the time cannot be negative, we take the positive solution.
Therefore, it takes approximately 0.947 seconds for the red ball to hit the ground.
To determine the maximum height the blue ball reaches, we can use the same equation of motion:
s = ut + (1/2)at^2
where:
s = displacement (the maximum height, relative to its start point, 1 meter)
u = initial velocity (23.5 m/s)
t = time (0.6 seconds)
a = acceleration due to gravity (-9.81 m/s^2, still negative)
Plugging in the values, we get:
1 = 23.5(0.6) + (1/2)(-9.81)(0.6)^2
Simplifying the equation, we have:
1 = 14.1 - 1.7661
1 ≈ 12.3339
Therefore, the maximum height the blue ball reaches is approximately 12.3339 meters above its starting point.